I am trying to understand implicit differentiation; I understand what to do (that is no problem), but why I do it is another story. For example: 3 y 2 = 5 x 3 I understand that, if I take the derivative with respect to x of both sides of the equation, I'll get: d d x ( 3 y 2 ) = d d x ( 5 x 3 ) 6 y d d x ( y ) = 15 x 2 d d x ( x ) 6 y d y d x = 15 x 2 d x d x 6 y d y d x = 15 x 2 d y d x = 15 x 2 6 y Unless I made some sort of error, this is what I am suppose to do. But why? Specifically, on the second line, I utilize the chain rule for the "outer function" and get 6y, but I still need to utilize the chain rule for the "inner function" which is the y. So why don't I go ahead and take the derivative of y and get 1? I know that I am not suppose to, but I don't really "get it." It seems to me that I only use the chain rule "halfway". Why isn't it an all or nothing? If it's all done with respect to x, it would seem to me that the 3y^2 should remain unchanged entirely. This is my problem.

Question

I am trying to understand implicit differentiation; I understand what to do (that is no problem), but why I do it is another story. For example:  3      y    2    =  5      x    3  I understand that, if I take the derivative with respect to x of both sides of the equation, I&#039;ll get:      d          d      x        (  3      y    2    )  =      d          d      x        (  5      x    3    )  6  y      d          d      x        (  y  )  =  15      x    2        d          d      x        (  x  )  6  y            d      y              d      x        =  15      x    2              d      x              d      x        6  y            d      y              d      x        =  15      x    2              d      y              d      x        =            15              x        2                    6      y      Unless I made some sort of error, this is what I am suppose to do. But why? Specifically, on the second line, I utilize the chain rule for the &quot;outer function&quot; and get 6y, but I still need to utilize the chain rule for the &quot;inner function&quot; which is the y. So why don&#039;t I go ahead and take the derivative of y and get 1? I know that I am not suppose to, but I don&#039;t really &quot;get it.&quot; It seems to me that I only use the chain rule &quot;halfway&quot;. Why isn&#039;t it an all or nothing? If it&#039;s all done with respect to x, it would seem to me that the 3y^2 should remain unchanged entirely. This is my problem.

Shane Middleton · Accepted Answer

d            y                      d            x       may equal 1 in special cases, but it is not generally so.   The very point of implicit derivation is that you don&#039;t know what it is.So, the Chain Rule is used to separate the derivative of a function of   y wrt   x into two differential terms, one you can resolve, and one unknown.To be clear, here&#039;s the process with the stage added and highlighted. Does this help?                    3                  y          2                                    =        5                  x          3                                                                        d                        (            3                          y              2                        )                                              d                        x                                              =                                            d                        (            5                          x              3                        )                                              d                        x                                      Take the derivative w.r.t.         x                                                                                d                            (              3                              y                2                            )                                                      d                            y                                                                          d                            y                                                      d                            x                                                          =                                            d                        (            5                          x              3                        )                                              d                        x                                      Apply the Chain Rule to the L.H.S.                            6        y                                            d                        y                                              d                        x                                              =        15                  x          2                            Evaluate the polynomial derivatives                                             d                        (            c                                      z              n                        )                                              d                        z                          =        c                n                          z                      n            −            1                                                          where         n        ∈                  N                ,         and         c         is constant                                                                d                        y                                              d                        x                                              =                              5                          x              2                                            2            y                                      Use arthimetic rearrangement            Alternatively, we might use explicit differentiation, as follows:                    3                  y          2                                    =        5                  x          3                                    y                            =                  x                      3                          /                        2                                                        5            3                                                ⋆                                                                d                        y                                              d                        x                                              =                                            d                                      x                              3                                  /                                2                                                                        d                        x                                                        5            3                                                                                     =                              3                          x                              1                                  /                                2                                                    5                                            2                          3                                                                                       =                              3                          x                              1                                  /                                2                                                    5                                            2                          3                                      ×                                            x                              3                                  /                                2                                                    5                                            y                          3                                                   re:         ⋆                                                         =                              5                          x              2                                            2            y

Sanai Ball · Accepted Answer

Change   y to   y  (  x  ) and everything should be clear.   y is not a constant, but a function of   x.            d      y              d      x        =      d          d      x        y  (  x  )  =      y    ′    (  x  )

I am trying to understand implicit differentiation; I understand what to do (that is no problem), but why I do it is another story. For example: 3y^2=5x^3

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